The question is: Prove algebraically that the difference between the squares of any two
consecutive even numbers is always a multiple of 4.
My working was: Let the numbers be n and n + 2
(n+2)^2 - (n)^2
=n^2 + 4n + 4 - n^2
=4n + 4
=4(n + 1)
But the answers say the numbers should be 2n + 2 and 2n, why is this or is mine right too?
Thanks
any even number is a multiple of 2.
When you pick any old n, it might not be even.
You have proven that any two numbers which differ by 2 have their squares differing by a multiple of 4.
So, technically you have not done what was asked, but have proved an even stronger result.
Using 2n, the proof is even easier:
(2n+2)^2 - (2n)^2
= 4(n+1)^2 - 4n^2
= 4((n+1)^2 - n^2)
QED
I understand thanks!
Your working is absolutely correct. The numbers can indeed be represented as n and n + 2, where n is any even number.
The general approach to solving this problem is to use algebraic manipulation to express the difference between the squares of these consecutive even numbers and then prove that it is always a multiple of 4.
Let's go through the steps of your solution:
1. Let the numbers be n and n + 2.
2. Express the difference between their squares: (n + 2)^2 - n^2.
3. Simplify the expression: n^2 + 4n + 4 - n^2.
4. Further simplify: 4n + 4.
5. Factor out 4: 4(n + 1).
As you correctly derived, the difference between the squares of any two consecutive even numbers is indeed a multiple of 4. So, your solution is absolutely correct.
Regarding the alternative representation of the numbers as 2n and 2n + 2, it is just another valid way to express consecutive even numbers. Both representations are correct, and the final result will be the same.
In conclusion, your solution is right, and both representations of the numbers are acceptable.